2 added 27 characters in body edited Feb 14 '15 at 18:23 bon 12.7k1010 gold badges4343 silver badges8080 bronze badges You don't need to add water to produce the amine, but it's the simplest way of destroying any remaining hydride reducing agent (to give aluminium hydroxide, lithium hydroxide and hydrogen). Presuming you're doing the reaction in ether, the LiOH$$\ce{LiOH}$$ will all go to the aqueous layer, whereas Al(OH)3$$\ce{Al(OH)3}$$ is pretty much insoluble in any solvent you're likely to use in organic synthesis, so it's easy to extract the amine from the aqeuous solvent. The "chemguide" page is a bit confusing. The reaction they write out isn't a great description of what's happening to the oxygen in the amide (it's probably going to be incorporated into Al(OH)3)$$\ce{Al(OH)3}$$ but the other two "[H]"s"$$\ce{[H]}$$"s just go to making the carbonyl oxygen into water. You don't need to add water to produce the amine, but it's the simplest way of destroying any remaining hydride reducing agent (to give aluminium hydroxide, lithium hydroxide and hydrogen). Presuming you're doing the reaction in ether, the LiOH will all go to the aqueous layer, whereas Al(OH)3 is pretty much insoluble in any solvent you're likely to use in organic synthesis, so it's easy to extract the amine from the aqeuous solvent. The "chemguide" page is a bit confusing. The reaction they write out isn't a great description of what's happening to the oxygen in the amide (it's probably going to be incorporated into Al(OH)3) but the other two "[H]"s just go to making the carbonyl oxygen into water. You don't need to add water to produce the amine, but it's the simplest way of destroying any remaining hydride reducing agent (to give aluminium hydroxide, lithium hydroxide and hydrogen). Presuming you're doing the reaction in ether, the $$\ce{LiOH}$$ will all go to the aqueous layer, whereas $$\ce{Al(OH)3}$$ is pretty much insoluble in any solvent you're likely to use in organic synthesis, so it's easy to extract the amine from the aqeuous solvent. The "chemguide" page is a bit confusing. The reaction they write out isn't a great description of what's happening to the oxygen in the amide (it's probably going to be incorporated into $$\ce{Al(OH)3}$$ but the other two "$$\ce{[H]}$$"s just go to making the carbonyl oxygen into water. 1 answered Feb 14 '15 at 18:21 J. LS 1,4451212 silver badges4343 bronze badges You don't need to add water to produce the amine, but it's the simplest way of destroying any remaining hydride reducing agent (to give aluminium hydroxide, lithium hydroxide and hydrogen). Presuming you're doing the reaction in ether, the LiOH will all go to the aqueous layer, whereas Al(OH)3 is pretty much insoluble in any solvent you're likely to use in organic synthesis, so it's easy to extract the amine from the aqeuous solvent. The "chemguide" page is a bit confusing. The reaction they write out isn't a great description of what's happening to the oxygen in the amide (it's probably going to be incorporated into Al(OH)3) but the other two "[H]"s just go to making the carbonyl oxygen into water.